\nLet the total work be 1 unit.
\nA can finish the work in 20 days, so A’s daily work rate is $\\frac{1}{20}$.
\nB can finish the work in 25 days, so B’s daily work rate is $\\frac{1}{25}$.
\nC can finish the work in 30 days, so C’s daily work rate is $\\frac{1}{30}$.
\nThey start working together (A, B, and C). Their combined daily work rate is $\\frac{1}{20} + \\frac{1}{25} + \\frac{1}{30}$.
\nTo add these fractions, find a common denominator (LCM of 20, 25, 30 is 300):
\nCombined rate = $\\frac{15}{300} + \\frac{12}{300} + \\frac{10}{300} = \\frac{15+12+10}{300} = \\frac{37}{300}$.
\nThey work together for 3 days. Work done in the first 3 days = $3 \\times \\frac{37}{300} = \\frac{37}{100}$.
\nAfter 3 days, B quits. The remaining work is $1 – \\frac{37}{100} = \\frac{63}{100}$.
\nThe remaining work is done by A and C. Their combined daily work rate is $\\frac{1}{20} + \\frac{1}{30}$.
\nLCM of 20 and 30 is 60:
\nCombined rate of A and C = $\\frac{3}{60} + \\frac{2}{60} = \\frac{5}{60} = \\frac{1}{12}$.
\nTime taken by A and C to finish the remaining work = $\\frac{\\text{Remaining Work}}{\\text{Combined Rate of A and C}} = \\frac{63\/100}{1\/12}$.
\nTime taken = $\\frac{63}{100} \\times 12 = \\frac{63 \\times 3}{25} = \\frac{189}{25}$ days.
\nThe question asks for the total number of days from the start.
\nTotal time = Time A, B, C worked together + Time A and C worked together.
\nTotal time = 3 days + $\\frac{189}{25}$ days.
\nTotal time = $\\frac{3 \\times 25}{25} + \\frac{189}{25} = \\frac{75+189}{25} = \\frac{264}{25}$ days.
\nConverting the improper fraction to a mixed number: $264 \\div 25$. $264 = 10 \\times 25 + 14$.
\nSo, $\\frac{264}{25} = 10\\frac{14}{25}$ days.
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